We have,


We know that sum of probability $=1$.
$$
\begin{aligned}
& \Sigma p\left(X=x_i\right)=1 \\
& \begin{array}{l}
\Rightarrow \quad \frac{1}{6}+k+\frac{1}{4}+k+\frac{1}{6}=1 \\
\Rightarrow \quad 2 k=1-\frac{7}{12} \Rightarrow k=\frac{5}{24}
\end{array} \\
& \text { Now, } E(X)=\Sigma x_i p_i \\
& \quad=-2\left(\frac{1}{6}\right)-1\left(\frac{5}{24}\right)+0\left(\frac{1}{4}\right)+1\left(\frac{5}{24}\right)+2\left(\frac{1}{6}\right) \\
& \begin{array}{c}
E(x)=0 \\
\text { Now, } E\left(X^2\right)=\Sigma x_i^2 p i \\
\quad=(-2)^2\left(\frac{1}{6}\right)+(-1)^2\left(\frac{5}{24}\right)+0\left(\frac{1}{4}\right) \\
\quad=\frac{4}{6}+\frac{5}{24}+\frac{5}{24}+\frac{4}{6}=\frac{16+10+16}{24}=\frac{42}{24}=\frac{7}{4}
\end{array} \\
& \begin{array}{l}
\operatorname{var}(x)=E(x)^2-[E(x)]^2 \\
\operatorname{var}(x)=\frac{7}{4}-0=\frac{7}{4}
\end{array}
\end{aligned}
$$